Solved Examples (Dedicated Problem Set)

Dedicated Problem Set — JEE/NEET Level

This is a curated set of 32 fully worked, exam-level problems covering every theme of Amines — classifying and naming, predicting products of ammonolysis, Gabriel and Hofmann bromamide routes, ordering basicity (aqueous and gas phase, aliphatic vs aromatic, substituted anilines), the carbylamine/Hinsberg/nitrous-acid distinctions, aniline electrophilic substitution, and the full range of diazonium reactions. Each solution shows the reasoning step by step. Work each one before reading the solution.

Naming, Classification & Basicity

Example 1. Classify and name (CH$_3$)$_2$CH-NH$_2$.

Solution: One carbon group on N → primary (1°) amine. IUPAC name propan-2-amine (isopropylamine).

Example 2. Arrange in increasing basicity (aqueous): NH$_3$, CH$_3$NH$_2$, (CH$_3$)$_2$NH, (CH$_3$)$_3$N.

Solution: NH$_3$ < (CH$_3$)$_3$N < CH$_3$NH$_2$ < (CH$_3$)$_2$NH. In water, solvation and steric effects make the 2° amine most basic and drop the 3° amine below the 1°.

Example 3. Why is aniline a weaker base than cyclohexylamine?

Solution: In aniline the N lone pair is delocalised into the benzene ring, so it is much less available to bind a proton. Cyclohexylamine is an ordinary aliphatic amine with a fully available lone pair, so it is far more basic.

Example 4. Arrange in increasing basicity: aniline, p-nitroaniline, p-methylaniline.

Solution: p-nitroaniline < aniline < p-methylaniline. -NO$_2$ (EWG) lowers basicity; -CH$_3$ (EDG) raises it.

Preparation — Carbon Count Traps

Example 5. Convert propanamide (CH$_3$CH$_2$CONH$_2$) to ethanamine.

Solution: Use Hofmann bromamide degradation (Br$_2$/NaOH); it removes one carbon, giving ethanamine (C$_2$H$_5$NH$_2$) from the 3-carbon amide.

Example 6. Convert CH$_3$CH$_2$Br to propan-1-amine (add a carbon).

Solution: CH$_3$CH$_2$Br + KCN → CH$_3$CH$_2$CN, then reduce (H$_2$/Ni or LiAlH$_4$) → CH$_3$CH$_2$CH$_2$NH$_2$ (propan-1-amine). The nitrile route adds one carbon.

Example 7. Which method gives a pure primary amine, and why?

Solution: The Gabriel phthalimide synthesis — the phthalimide nitrogen bears only one R group, so hydrolysis releases only the 1° amine (no 2°/3° contamination). Note: it works only for aliphatic primary amines.

Example 8. Why can't aniline be made by the Gabriel synthesis?

Solution: It would need an aryl halide to alkylate potassium phthalimide, but aryl halides do not undergo this SN2 reaction, so the aromatic amine cannot be made this way.

Distinguishing Tests

Example 9. Distinguish ethylamine, diethylamine and triethylamine by the Hinsberg test.

Solution: With benzenesulphonyl chloride then KOH: ethylamine (1°) → KOH-soluble product; diethylamine (2°) → insoluble precipitate; triethylamine (3°) → no reaction.

Example 10. Which amine gives a foul smell with CHCl$_3$ and alcoholic KOH?

Solution: A primary amine (the carbylamine/isocyanide test is positive for 1° amines only).

Example 11. How do you distinguish ethylamine from aniline using nitrous acid?

Solution: With cold HNO$_2$: ethylamine releases N$_2$ gas (gives ethanol); aniline forms a stable diazonium salt that couples (e.g. with 2-naphthol) to a red/orange dye.

Example 12. What does a secondary amine give with nitrous acid?

Solution: A yellow oily N-nitrosamine (R$_2$N-N=O).

Aniline Substitution

Example 13. Give the product of aniline + bromine water.

Solution: A white precipitate of 2,4,6-tribromoaniline (the strongly activated ring is trisubstituted).

Example 14. Why is aniline acetylated before nitration?

Solution: The acidic nitrating mixture would protonate -NH$_2$ to -NH$_3^+$ (meta-directing). Acetylation gives -NHCOCH$_3$, which stays o/p-directing and less reactive, so nitration gives mainly the para product (→ p-nitroaniline after hydrolysis).

Example 15. Why does aniline not undergo Friedel-Crafts alkylation?

Solution: AlCl$_3$ bonds to the basic nitrogen lone pair, giving N a positive charge that strongly deactivates the ring (meta-directing), so the reaction fails.

Example 16. Predict the product of sulphonation of aniline.

Solution: Sulphanilic acid (p-aminobenzenesulphonic acid), which exists as a zwitterion.

Diazonium Replacement Reactions

Example 17. How is chlorobenzene prepared from aniline?

Solution: Diazotise (NaNO$_2$/HCl, 273-278 K) to C$_6$H$_5$N$_2^+$Cl⁻, then CuCl (Sandmeyer) → chlorobenzene + N$_2$.

Example 18. Convert aniline to iodobenzene.

Solution: Diazotise, then warm with KI → iodobenzene (no copper catalyst needed).

Example 19. How is benzonitrile made from a diazonium salt?

Solution: Treat with CuCN (Sandmeyer) → benzonitrile (C$_6$H$_5$CN).

Example 20. How can the -NH$_2$ group be removed from aniline entirely (replaced by H)?

Solution: Diazotise, then treat with hypophosphorous acid (H$_3$PO$_2$) → benzene (the -N$_2^+$ is replaced by -H).

Diazonium Coupling & Multistep

Example 21. What forms when benzene diazonium chloride couples with phenol?

Solution: p-Hydroxyazobenzene (an orange azo dye), formed by para coupling through the -N=N- bridge.

Example 22. Name the dye formed by coupling benzene diazonium chloride with aniline.

Solution: p-Aminoazobenzene (aniline yellow).

Example 23. Convert benzene to p-bromoaniline (multistep).

Solution: Nitrate benzene → nitrobenzene; reduce → aniline; acetylate (protect) → acetanilide; brominate → p-bromoacetanilide; hydrolyse → p-bromoaniline. (Protection ensures mono-para bromination.)

Example 24. Convert aniline to phenol.

Solution: Diazotise (cold), then warm with water (H$_2$O/H⁺) → phenol + N$_2$.

Mixed JEE/NEET Challenge

Example 25. An organic compound C$_2$H$_7$N gives a foul smell with CHCl$_3$/KOH and, with HNO$_2$, releases N$_2$. Identify it.

Solution: Foul carbylamine smell ⇒ primary amine; N$_2$ with HNO$_2$ ⇒ aliphatic 1° amine. C$_2$H$_7$N primary aliphatic amine = ethanamine (CH$_3$CH$_2$NH$_2$).

Example 26. A C$_7$H$_9$N compound is a primary aromatic amine, slightly more basic than aniline, and gives p-substituted products. Suggest a structure.

Solution: A methyl-substituted aniline, e.g. p-toluidine (4-methylaniline) — the -CH$_3$ (EDG) makes it slightly more basic than aniline.

Example 27. Why is the gas-phase basicity order of amines different from the aqueous order?

Solution: In the gas phase only the electron-donating +I effect operates (3° > 2° > 1° > NH$_3$). In water, the solvation of the cation (better for cations with more N-H) and steric hindrance also matter, giving the irregular aqueous order.

Example 28. An amine does not react in the Hinsberg test (no precipitate, no soluble product). What class is it?

Solution: A tertiary (3°) amine — it has no N-H to react with benzenesulphonyl chloride.

Example 29. Convert aniline to 1,3,5-tribromobenzene.

Solution: Brominate aniline (bromine water) → 2,4,6-tribromoaniline; diazotise, then treat with H$_3$PO$_2$ to replace -N$_2^+$ by -H → 1,3,5-tribromobenzene.

Example 30. Which is more basic in water, ammonia or aniline, and why?

Solution: Ammonia. Its lone pair is fully available; aniline's lone pair is delocalised into the ring, so ammonia is the stronger base.

Example 31. Identify the reaction: C$_6$H$_5$N$_2^+$Cl⁻ + CuBr → C$_6$H$_5$Br + N$_2$.

Solution: The Sandmeyer reaction (replacement of -N$_2^+$ by -Br using cuprous bromide).

Example 32. A primary aromatic amine is treated with NaNO$_2$/HCl at 5 °C and the product is added to alkaline 2-naphthol, giving a bright red precipitate. What does this confirm?

Solution: The formation of an azo dye confirms a primary aromatic amine (it formed a diazonium salt that coupled with 2-naphthol).

JEE Main & Advanced Level Solved Examples

These problems push into mechanism, basicity reasoning, selective synthesis and the diazonium toolkit — the discriminators that decide JEE ranks. Decide the class of amine (1°/2°/3°, aliphatic/aromatic) and whether the nitrogen lone pair is available before you answer.

Example 33: Hofmann elimination (exhaustive methylation) [JEE Advanced]

When sec-butyltrimethylammonium hydroxide is heated, which alkene is the major product, and how does this differ from acid-catalysed dehydration?

Solution: Heating a quaternary ammonium hydroxide gives an alkene by Hofmann elimination. The leaving group (-N$^+$(CH$_3$)$_3$) is bulky, so the base removes the most accessible (least hindered) beta-hydrogen, giving the least substituted alkene (the Hofmann product) — here but-1-ene predominates over but-2-ene. This is the opposite of Saytzeff selectivity seen in acid-catalysed alcohol dehydration, where the more substituted alkene wins.

Example 34: Mechanism of Hofmann bromamide degradation [JEE Advanced]

An amide RCONH$_2$ treated with Br$_2$ and NaOH gives a primary amine with one fewer carbon. Explain the mechanism and the stereochemical outcome.

Solution: The steps are: (i) base removes an N-H and Br$_2$ gives the N-bromoamide; (ii) a second deprotonation gives a bromamide anion that loses Br$^-$ to form an electron-deficient nitrene-like species; (iii) the R group migrates from carbon to nitrogen (a 1,2-shift) to give an isocyanate, R-N=C=O; (iv) alkaline hydrolysis of the isocyanate loses CO$_2$ to give R-NH$_2$. The migrating group keeps its bond pair and moves with retention of configuration, and because the carbonyl carbon is lost as carbonate, the amine has one fewer carbon than the amide.

JEE/Advanced — Basicity & Separation

Example 35: Relating pK$_b$ and pK$_a$ [JEE]

The pK$_b$ of methylamine is 3.36. What is the pK$_a$ of its conjugate acid (the methylammonium ion), and what does the value indicate?

Solution: For a base and its conjugate acid in water at 25 °C, K$_a$ x K$_b$ = K$_w$, so pK$_a$ + pK$_b$ = 14. Hence pK$_a$(CH$_3$NH$_3^+$) = 14 - 3.36 = 10.64. A high pK$_a$ of the conjugate acid means the methylammonium ion holds its proton tightly, i.e. methylamine is a relatively strong base — stronger than ammonia (whose conjugate acid has pK$_a$ 9.25).

Example 36: Basicity of nitrogen heterocycles [JEE Advanced]

Arrange pyrrole, pyridine, aniline and ammonia in increasing order of basicity, and explain the position of pyrrole and pyridine.

Solution: Basicity depends on how available the nitrogen lone pair is.

• In pyrrole the lone pair is part of the aromatic 6 pi-electron sextet; donating it would destroy aromaticity, so pyrrole is the weakest base.
• In aniline the lone pair is partly delocalised into the benzene ring — weakly basic.
• In pyridine the lone pair sits in an sp$^2$ orbital in the ring plane, outside the aromatic system, so it is freely available — more basic than aniline.
• Ammonia has a fully available lone pair and no competing delocalisation.

Order: pyrrole < aniline < pyridine < ammonia (aliphatic amines such as methylamine are stronger still).

Example 37: Separating a mixture of 1°, 2° and 3° amines (Hinsberg method) [JEE]

How does the Hinsberg reagent allow a mixture of primary, secondary and tertiary amines to be separated?

Solution: Hinsberg reagent is benzenesulphonyl chloride (C$_6$H$_5$SO$_2$Cl).

• A 1° amine gives an N-substituted sulphonamide that still has an acidic N-H, so it dissolves in KOH (alkali-soluble).
• A 2° amine gives a sulphonamide with no N-H, so it is insoluble in KOH (precipitates).
• A 3° amine has no N-H to react and does not form a sulphonamide — it stays as the free amine and can be extracted/distilled off.

So treating the mixture with the reagent and KOH separates all three; acidifying the alkaline solution then regenerates the 1° amine.

JEE/Advanced — Reactions with Nitrous Acid & Reductive Amination

Example 38: Tertiary aromatic amine with nitrous acid [JEE Advanced]

N,N-dimethylaniline reacts with cold HNO$_2$ to give a green solid, whereas trimethylamine merely forms a soluble salt. Explain.

Solution: A tertiary aliphatic amine (e.g. (CH$_3$)$_3$N) has no N-H and no activated ring, so HNO$_2$ only forms a water-soluble nitrite salt. In N,N-dimethylaniline the ring is strongly activated by the -N(CH$_3$)$_2$ group, so the nitrosonium ion (NO$^+$) attacks the ring in electrophilic C-nitrosation at the para position, giving p-nitroso-N,N-dimethylaniline (a green solid). So 3° aromatic amines undergo ring nitrosation, unlike 3° aliphatic amines.

Example 39: Reductive amination for a clean secondary amine [JEE]

Direct alkylation of ammonia gives a mixture of 1°, 2°, 3° amines and a quaternary salt. How can N-methylethanamine (a single 2° amine) be made selectively?

Solution: Use reductive amination. Condense the carbonyl with the amine to form an imine, then reduce it: CH$_3$CHO + CH$_3$NH$_2$ → imine (CH$_3$CH=N-CH$_3$) → (H$_2$/Ni or NaBH$_3$CN) → CH$_3$CH$_2$-NH-CH$_3$ (N-methylethanamine). Because the imine is reduced to a single 2° amine, the over-alkylation mixture of direct ammonolysis is avoided.

JEE/Advanced — Diazonium Salts in Synthesis

Example 40: pH control in azo coupling [JEE Advanced]

Benzene diazonium chloride couples with phenol in mildly alkaline medium but with aniline in mildly acidic medium. Explain the choice of pH in each case.

Solution: The diazonium ion is a weak electrophile, so the partner must be electron-rich, but the medium must not destroy the diazonium salt. With phenol: mild alkali converts it to the far more nucleophilic phenoxide ion, speeding coupling; strong alkali is avoided because it would convert the diazonium ion to an unreactive diazotate. With aniline: a mildly acidic/neutral medium keeps enough free amine (a good nucleophile) available — too much acid would fully protonate the -NH$_2$ (deactivating it), while alkalinity is unnecessary. The azo product (e.g. p-hydroxyazobenzene) forms at the para position.

Example 41: Why fluorobenzene needs the Balz-Schiemann reaction [JEE Advanced]

Chloro-, bromo- and iodobenzene are made from benzene diazonium salts by the Sandmeyer/Gattermann routes, but fluorobenzene is not. How is it made?

Solution: Fluorobenzene is prepared by the Balz-Schiemann reaction. The diazonium chloride is treated with fluoroboric acid (HBF$_4$) to precipitate the diazonium fluoroborate (Ar-N$_2^+$BF$_4^-$); heating the dry salt then gives Ar-F + N$_2$ + BF$_3$. The Cu(I)-halide (Sandmeyer) route does not deliver fluoride, so this thermal decomposition is the standard way to put fluorine on the ring.

Example 42: Multi-step synthesis of 1,3,5-tribromobenzene [JEE Advanced]

Direct bromination of benzene cannot give 1,3,5-tribromobenzene. Devise a synthesis from benzene.

Solution: Exploit the strongly o/p-directing -NH$_2$ group, then remove it via a diazonium salt:

1. Benzene → nitrobenzene (conc. HNO$_3$/H$_2$SO$_4$) → aniline (Sn/HCl or H$_2$/Ni reduction).
2. Aniline + 3 Br$_2$ (bromine water) → 2,4,6-tribromoaniline (the -NH$_2$ directs Br to all three o/p positions).
3. Diazotise (NaNO$_2$/HCl, 273-278 K), then deaminate by warming with hypophosphorous acid (H$_3$PO$_2$), which replaces the diazonium group by -H → 1,3,5-tribromobenzene.

The amino group acts as a removable director that places the three bromines in the otherwise inaccessible 1,3,5 pattern.